Question

What is the area of a figure formed by a square of side 8 cm and an isosceles triangle with the base as one side of the square and the other equal sides as 5 cm?

• A. $70c{m}^2$
• B. $76c{m}^2$
• C. $82c{m}^2$
• D. $64c{m}^2$

Answer 1

The correct option is B $76c{m}^2$

Each of the equal sides of the isosceles $\Delta EAB=\frac{18-8}{2}cm=5cm$

$\therefore$ In the isosceles $\Delta EAB$, altitude $EF$ bisects the base $AB$

$\Rightarrow AF=FB=4cm$

In $\Delta EFA$,

$EF=\sqrt{E{A}^2-A{F}^2}=\sqrt{25-16}=\sqrt{9}cm=3cm$

Area of the figure $=$ Area of $\Delta EAB+$ Area of square $ABCD$

$=\frac{1}{2}\times 8\times 3c{m}^2+(8\times 8)c{m}^2$

$=12+64=76c{m}^2$